The Mandelstam Variable s
Introduce the Mandelstam variable $s=(p_{1}+p_{2})^{2}$ and show that
$4\left[(p_{1}\cdot p_{2})^{2}-m_{1}^{2}m_{2}^{2}\right]=\left[s-(m_{1}+m_{2})^{2}\right]\left[s-(m_{1}-m_{2})^{2}\right]$
holds.

Question

The Mandelstam Variable s
Introduce the Mandelstam variable [latex]s=(p_{1}+p_{2})^{2}[/latex] and show that

[latex]4\left[(p_{1}\cdot p_{2})^{2}-m_{1}^{2}m_{2}^{2}\right]=\left[s-(m_{1}+m_{2})^{2}\right]\left[s-(m_{1}-m_{2})^{2}\right][/latex]

holds.

Accepted Answer

The Mandelstam variable s provides an invariant measure for the energy of the particles participating in the reaction. In the center-of-mass system, where one has [latex]p_{1}+p_{2}=0,\,{\sqrt{s}}[/latex] is equal to the sum of all particle energies:

[latex]s=(p_{1}+p_{2})^{2}=(E_{1}+E_{1},\;p_{1}+p_{2})^{2}=(E_{1}+E_{2})^{2}\;.[/latex]

In general

[latex]s=(p_{1}+p_{2})^{2}=p_{1}^{2}+2\,p_{1}\cdot p_{2}+p_{2}^{2}[/latex]

[latex]=m_{1}^{2}+2\,p_{1}\cdot p_{2}+m_{2}^{2}\,\,,[/latex]

i.e.,

[latex]2\;p_{1}\cdot p_{2}=s-m_{1}^{2}-m_{2}^{2}\ .[/latex] (1)

The flux factor [latex]4\left((p_{1}\cdot p_{2})^{2}-m_{1}^{2}m_{2}^{2}\right)[/latex] then assumes the form

[latex](2\,p_{1}\cdot p_{2}-2m_{1}m_{2})(2\,p_{1}\cdot p_{2}+2m_{1}m_{2})[/latex]

[latex]=\left[s-(m_{1}+m_{2})^{2}\right]\left[s-(m_{1}-m_{2})^{2}\right]~.[/latex]

Question 2.3: The Mandelstam Variable s Introduce the Mandelstam variable ......

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